Calculating Possibilities for Winning Tickets | Combinations Problem

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To determine the number of possibilities of having at least one winning ticket when buying 6 out of 10 tickets, where 3 are winning, the calculation involves combinations. The total combinations of choosing 6 tickets from 10 is represented as C(10, 6). To find the scenarios with at least one winning ticket, subtract the combinations of selecting only losing tickets (C(7, 6)) from the total combinations. The correct approach is to calculate C(10, 6) - C(7, 6), ensuring the focus remains on the "at least" requirement. This method effectively captures the desired outcome of having at least one winning ticket.
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Homework Statement



From 10 tickets of luck, only 3 are winning. If we buy 6 tickets, how many possibilities we have at least to have 1 winning ticket in our hand?

Homework Equations



C_n^k=\frac{n!}{k!(n-k)!}

The Attempt at a Solution



6 tickets from 10, we can choose on C_1_0^6. There are 4 tickets left. C_7^3[/tex] is the tickets which are not winning. Is the right answer:C_1_0^6 - C_7^3<br /> <br /> ?
 
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Focus on the "at least" part.
 
Sorry, but I can't figure out
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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